Question

During chemical reaction 7.55gKI and 9.06g were allowed to react. How many grams of excess reagent are left over after the reaction is complete. Reaction: Pb(NO3)2(s) + 2KCI(s) > 2KNO3(s) + PbI(s)

Answer 1

Answer: 7.45 g of
`Pb(NO_3)_2` excess reagent are left over after the reaction is complete.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

a)
`{\text{Number of moles of} KI}=(7.55g)/(166g/mol)=0.045moles`

b)
`{\text{Number of moles of} Pb(NO_3)_2}=(9.06g)/(331.2g/mol)=0.027moles`

The balanced chemical reaction is :

`Pb(NO_3)_2(s)+2KI(s)\rightarrow 2KNO_3(s)+PbI(s)`

According to stoichiometry :

2 moles of
`KI` require = 1 mole of
`Pb(NO_3)_2`

Thus 0.045 moles of
`KI` will require=
`(1)/(2)* 0.045=0.0225moles` of
`Pb(NO_3)_2`

Thus
`KI` is the limiting reagent as it limits the formation of product and
`Pb(NO_3)_2` is the excess reagent as (0.045-0.0225) = 0.0225 moles are left

Mass of
`Pb(NO_3)_2=moles* {\text {Molar mass}}=0.0225moles* 331.2g/mol=7.45g`

Thus 7.45 g of
`Pb(NO_3)_2` of excess reagent are left over after the reaction is complete.